1) Each of the following would be considered company-confidential except

A man weighs 145 lb on earth.Part ASpecify his mass in slugs.Express your answer to three significant figures and include the ap

adell [148]

Answer:

a) 4.51 lbf-s^2/ft

b) 65.8 kg

c) 645 N

d) 23.8 lb

e) 65.8 kg

Explanation:

Weight of the man on Earth = 145 lb

a) Mass in slug is...

32.174 pound = 1 slug

145 pound = $x$ slug

$x$ = 145/32.174 = 4.51 lbf-s^2/ft

b) Mass in kg is...

2.205 pounds = 1 kg

145 pounds = $x$ kg

$x$ = 145/2.205 = 65.8 kg

c) Weight in Newton = mg

where

m is mass in kg

g is acceleration due to gravity on Earth = 9.81 m/s^2

Weight in Newton = 65.8 x 9.81 = 645 N

d) If on the moon with acceleration due to gravity of 5.30 ft/s^2,

1 m/s^2 = 3.2808 ft/s^2

$x$ m/s^2 = 5.30 ft/s^2

$x$ = 5.30/3.2808 = 1.6155 m/s^2

weight in Newton = mg = 65.8 x 1.6155 = 106

weight in pounds = 106/4.448 = 23.8 lb

e) The mass of the man does not change on the moon. It will therefore have the same value as his mass here on Earth

mass on the moon = 65.8 kg

3 0

3 years ago

The convection heat transfer coefficient for a clothed person standing in moving air is expressed as h 5 14.8V0.69 for 0.15 , V

Rom4ik [11]

Cychbjnivrxezyyihvhuytrruokjaa

7 0

3 years ago

Read 2 more answers

An air compressor of mass 120 kg is mounted on an elastic foundation. It has been observed that, when a harmonic force of amplit

kupik [55]

Answer:

equivalent stiffness is 136906.78 N/m

damping constant is 718.96 N.s/m

Explanation:

given data

mass = 120 kg

amplitude = 120 N

frequency = 320 r/min

displacement = 5 mm

to find out

equivalent stiffness and damping

solution

we will apply here frequency formula that is

frequency ω = ω(n) √(1-∈ ²) ......................1

here ω(n) is natural frequency i.e = √(k/m)

so from equation 1

320×2π/60 = √(k/120) × √(1-2∈²)

k × ( 1 - 2∈²) = 33.51² ×120

k × ( 1 - 2∈²) = 134752.99 .....................2

and here amplitude ( max ) of displacement is express as

displacement = force / k × ( $\frac{1}{2\varepsilon \sqrt{1-\varepsilon ^2}}$ )

put here value

0.005 = 120/k × ( $\frac{1}{2\varepsilon \sqrt{1-\varepsilon ^2}}$ )

k ×∈ × √(1-2∈²) = 1200 ......................3

so by equation 3 and 2

$\frac{k\varepsilon \sqrt{1-\varepsilon^2})}{k(1-2\varepsilon^2)} = \frac{12000}{134752.99}$

$\varepsilon^{2} - \varepsilon^{4} = 7.929 * 10^{-3} - 0.01585 * \varepsilon^{2}$

solve it and we get

∈ = 1.00396

and

∈ = 0.08869

here small value we will consider so

by equation 2 we get

k × ( 1 - 2(0.08869)²) = 134752.99

k = 136906.78 N/m

so equivalent stiffness is 136906.78 N/m

and

damping is express as

damping = 2∈ √(mk)

put here all value

damping = 2(0.08869) √(120×136906.78)

so damping constant is 718.96 N.s/m

7 0

3 years ago

How old are legos? Who created them? Why did they create them?

Sunny_sXe [5.5K]

“How old are the legos” 89 years old.

“Who created them?” Ole Kirk Christiansen.

“Why did they created them?” As a you man, Christiansen turned his love of whittling and playing with wood into a Business and, in 1916, he opened his own shop. Since times were so hard, Christiansen made the hard decision to use his wood to create inexpensive goods that might actually sell. Among them were Cheap toys.

Hope this helps•

Bye~

•Kate•

4 0

3 years ago

If the rotational speed of a pump motor is reduced by 35%, what is the effect on the pump performance in terms of capacity, head

FinnZ [79.3K]

Answer:

- the capacity of the pump reduces by 35%.

- the head gets reduced by 57%.

the power consumption by the pump is reduced by 72%

Explanation:

the pump capacity is related to the speed as speed is reduces by 35%

so new speed is (100 - 35) = 65% of orginal speed

speed Q ∝ N ⇒ Q1/Q2 = N1/N2

Q2 = (N2/N1)Q1

Q2 = (65/100)Q1

which means that the capacity of the pump is also reduces by 35%.

the head in a pump is related by

H ∝ N² ⇒ H1/H2 = N1²/N2²

H2 = (N2N1)²H1

H2 = (65/100)²H1 = 0.4225H1

so the head gets reduced by 1 - 0.4225 = 0.5775 which is 57%.

Now The power requirement of a pump is related as

P ∝ N³ ⇒ P1/P2 = N1³/N2³

P2 = (N2/N1)³P1

H2 = (65/100)²P1 = 0.274P1

So the reduction in power is 1 - 0.274 = 0.725 which is 72%

Therefore for a reduction of 35% of speed there is a reduction of 72% of the power consumption by the pump.

8 0

3 years ago